Understanding Flight: Pressure Distribution & the Science Behind Airplanes

[arildno]

isnt it so that airplanes can fly upside down?
then this stuff with the wing pushing air down doesn't work really...

A plane can only fly upside down if it tilts its wing into such a position that it effectively produces downwash (turning the air downwards)
That this tilted wing-geometry is lift-sustaining, is encapsulated in the fact that although somewhat inverted, the wing's EFFECTIVE angle of attack remains positive..

 

[arildno]

Would it? Why do satellites then fall down after some time due to drag in the upper atmosphere? At a height of 500 km the density of air (mainly atomic oxygen) is about 10^8 cm^-3 which, assuming a collision cross section of 10^-16 cm^2, amounts to a free flight distance of 10^8 cm = 1000 km between collisions of two atoms. Surely more than enough to assume an inviscid gas.

I suggest you learn the difference between FLYING and FALLING before posting next time.

 

[russ_watters]

The one and only reason why airplanes fly is the fact the the orientation and shape of the wing in combination with the velocity relative to the air leads to more molecules hitting the wing from below than from above.

Ahh, see that clarifies something you said in your other thread - the thing about a flat-bottom wing. By that above logic, a flat bottom wing should produce negative lift at 0 aoa because there are no air particles hitting the bottom surface and a lot hitting the top. But you already know that isn't true: they produce lift even at a few degrees negative aoa.

You just disproved your own hypothesis.

isnt it so that airplanes can fly upside down?
then this stuff with the wing pushing air down doesn't work really...

High performance aircraft have symmetrical cross section wings so that they perform exactly the same whether right side up or upside down.

 

[FredGarvin]

Would it? Why do satellites then fall down after some time due to drag in the upper atmosphere? At a height of 500 km the density of air (mainly atomic oxygen) is about 10^8 cm^-3 which, assuming a collision cross section of 10^-16 cm^2, amounts to a free flight distance of 10^8 cm = 1000 km between collisions of two atoms.

I am in awe.

Surely more than enough to assume an inviscid gas.

The viscosity can infinitely approach zero, just not equal zero. No matter how small the viscosity, there will be a boundary layer and thus separation and thus form drag.

 

[Astronuc]

FoilSim II Version 1.5a - http://www.grc.nasa.gov/WWW/K-12/airplane/foil2.html

http://www.grc.nasa.gov/WWW/K-12/airplane/bga.html

http://www.grc.nasa.gov/WWW/K-12/airplane/presar.html


There are actually two modes of flight - 1) with flaps down and 2) trim with flaps up.

1) Flaps down when forward velocity cannot produce sufficient lift on wing foil, to flaps divert flow down. This occurs with take off and landing.

2) Trim with flaps up - forward speed produces lower pressure on top of wing (see FoilSim).

Some planes can fly upside down, provided they have a greater angle of attack, and thrust from the engine/propeller may play a role. This is confined to jet fighters and certain types of aircraft, e.g. many biplanes and acrobatic aircraft. Large aircraft do not fly upside down.

I was recently on a commercial flight and was aware that the airliner was flying with a greater than usual pitch.

 

[arildno]

The viscosity can infinitely approach zero, just not equal zero. No matter how small the viscosity, there will be a boundary layer and thus separation and thus form drag.

This is PRECISELY the issue here!
Thank you for emphasizing this.
The Euler equations can often be regarded as the leading order solution (for small viscosities) to the Navier-Stokes equations.
Unfortunately, the relation between E. and N-S is that we really have a SINGULAR perturbation problem, rather than a regular perturbation problem.

The flow as predicted by the Euler equations in the case of steady motion under stationary conditions with no initial circulation is completely, will be totally misleading if we proceed as if we had to do with a regular perturbation problem.

This does not deny the value of the Euler equations; it merely shows we need to proceed with extreme care as to determine when this set of equations yields immense benefits in the form of (sufficiently) accurate predictions&huge mathematical simplification, or when they will provide wildly inaccurate results.
When the Euler equations fails to work properly, viscosity is, in general, the culprit.

Astronuc: Thanks for the links&info.
I find glenn research centre's pages to be one of the best sites to explain important aspects of aerodynamics for a general public.

 

[PBRMEASAP]

There is a good topological argument for why a material curve should not usually get broken up:
Consider the positions of a material curve at times "t" and "t+dt".
Since the constituent particles have finite velocities, we should expect that we can map the "t" curve onto the "t+dt" curve through a CONTINUOUS transformation (that is, given "sufficient" closeness of points on the "t" curve, their images will be satisfactorily close on the "t+dt"-curve.)
But, can a continuous transformation effect the radical topological change from "closed" to "not closed" (think of the famous rubber band analogy of topology)?
This seems very unlikely; I am in fact, quite convinced it is untrue

You are right, a continuous transformation maps a closed curve to another closed curve. In the little book by Chorin and Marsden (Intro. to Mathematical Fluid Mech.), they call this the "fluid flow map". They even assume it is differentiable. Of course, I guess you can assume whatever you want when you develop a mathematical theory of something. Whether or not it corresponds to reality is for experiments to decide. But the fluid flow map seems very reasonable. I also don't know anything about the existence and uniqueness of solutions to IBV problems. My knowledge of differential equations basically consists of a bag o' tricks. If I can find the solution, then it probably exists .

Astronuc:
Those are some great links. I like how they give examples of incorrect theories and show in detail why they are wrong. Very good information. Thanks for posting 'em.

 

[arildno]

Note that in 3-D real life, this type of conundrums are solved in that the wing is perfectly able to shed off an unbroken material curve..

Thanks for the Chorin&Marsden reference; I'll check it out.

 

[arildno]

The counter-spinning vortex theory

Make no mistake:
If we start with an unbounded inviscid fluid at rest, and an object starts moving through it with constant velocity, then when stationary conditions ensues, D'Alembert's paradox rears its ugly head.

Nonetheless, there is a theory floating about which seeks to explain the generation of lift SOLELY WITHIN THE LIMITS OF POTENTIAL THEORY!
Since you unfortunately can find references to this faulty theory in quite advanced fluid mechanics texts, it is important to be armed against it.

Their "argument" is dreadfully simple:
Suppose you start with a uniform stream (no circulation here).
Assume that at a given point, you have a coincident placement of two point vortices of opposing circulation (that would initially sum up to no NET circulation and no generated velocity field from them)
Let the "counter-spinning" vortex start moving away from the other vortex (with, say, a constant velocity).
Now, since the effect of a 2-D point vortex decays as $$\frac{1}{r}$$ , where r is the distance to the vortex centre, then, as time goes by, the velocity field in an arbitrary vicinity of the remaining point vortex will look like the velocity field of a translatory potential plus that induced by the point vortex (which HAS non-zero circulation).
This could, for example, "explain" the Magnus effect (lift is essentially a WARPED Magnus effect).

Now, not commenting on the "physics" of spontaneously generated vortex pairs, it is simple enough to see why this theory is mathematically illucid as well:
While apparently a solution of the Laplace equation, even if you hide away the remaining vortex within a cylinder (as in the Magnus case), that counter-spinning vortex will enter the fluid domain AT A FINITE TIME. Since, however, our solution must be regular at all fluid points at all finite times, this shows the inherent worthlessness of the counter-spinning vortex "theory".
Another insoluble problem is, of course, the fulfillment of boundary conditions..

This post was just a warning to you of the nonsense which has sometimes been presented as science..

Of course, neither would it work to try and work out a point vortex theory with time-varying strengths.
Although these are valid solutions of Laplace's equation, they contradict Kelvin's theorem.
There simply don't exist the type of forces in inviscid, barotropic fluids which could account for the thereby induced velocity fields.

 

[Thomas2]

The viscosity can infinitely approach zero, just not equal zero. No matter how small the viscosity, there will be a boundary layer and thus separation and thus form drag.

Obviously, no physical quantity is exactly zero in the mathematical sense, but that doesn't mean one can't set them to zero in practice under certain circumstances. What you have to do here is to compare the size of the object to the mean free path between two collisions of molecules in the gas. If the latter is much larger than the former, one can certainly assume that the gas can be treated as inviscid (I mean what kind of boundary layer would you expect if the molecules don't collide with each other within 1000 km of the surface of the object?).

The flaw with the conclusions from the Potential Flow problem that lead to d'Alembert's paradox (see for instance http://astron.berkeley.edu/~jrg/ay202/node95.html ) is that the assumption of an inviscid potential flow is a contradiction in terms. In a strictly inviscid gas there is no interaction of molecules at all and the molecules just hit the object according to its geometrical cross section, with the rest of the gas stream completely unaffected by the object. Since the molecules hitting the object transfer momentum to it, there must hence also be a drag in an inviscid fluid.

 

[Thomas2]

Ahh, see that clarifies something you said in your other thread - the thing about a flat-bottom wing. By that above logic, a flat bottom wing should produce negative lift at 0 aoa because there are no air particles hitting the bottom surface and a lot hitting the top. But you already know that isn't true: they produce lift even at a few degrees negative aoa.
You just disproved your own hypothesis

Yes, if the upper surface of the airfoil is shaped such that the area exposed to the airstream (which causes a negative lift) is larger than the area in the shadow of the airstream (which causes a positive lift) then you would be right. But that's not how airfoils are designed as far as I am aware.

 

[russ_watters]

Yes, if the upper surface of the airfoil is shaped such that the area exposed to the airstream (which causes a negative lift) is larger than the area in the shadow of the airstream (which causes a positive lift) then you would be right. But that's not how airfoils are designed as far as I am aware.

Besides being generally wrong, you are also contradicting yourself. That's not what you said in the piece I quoted.

In fact, if the "shadow" of the airstream made a difference, then a flat-bottom airfoil at -4* aoa, or better yet, an airfoil with a large camber (concave underside), would produce a fair amount of negative lift, considering that the entire bottom is in the "shadow". Once again, what you describe does not fit reality. And your contradictions and permutations make it sound like you are making this stuff up as you go along. Not only have you not bothered to learn how it really works, you haven't even thought through your own idea.

 

[arildno]

The lift-generation process.

As we have seen, a viscous fluid will favour downwash above upflow, i.e, that is, the fluid will (just) be able to kick off a vortex from the upper surface, and push the stagnation point towards the trailing edge.
As the vortex is shedded, the downrushing fluid can be imagined to insert itself at that place, that is, there is an attachment of the inviscid fluid onto the surface, and by kicking off that vortex, it is reasonable to suppose that the local pressure there decreases (it is no longer a stagnation point)
Thus, as new particles comes rushing along downwards, they gain a velocity increase (relative to those which were there before) due to Bernoulli, i.e, circulation is increased in two ways: the path has lengthened, and the velocities increased.

The typical "punch" by which now the downrushing fluid meets the next formed vortex should therefore have been strengthened, that is, it should more easily/faster dislodge the new vortex.
That is, by the initial asymmetry we necessarily must have, we have entered a CASCADE process which might be imagined for example like this:
Transient separation/Vortex Formation->Vortex Shedding->Pressure Decrease/Attachment of streamline->Circulation increase->Meeting new stagnation point->back again.
Clearly, this cascade must eventually slow down, and an easy way to see this, is that the upper and lower fluid domains must merge/collide at the backside, i.e, a sufficiently strong stagnation pressure will develop at the trailing edge.
Therefore, the process will slow down after a while, and the final circulation&lift value is reached.

Thus, if this picture is roughly correct, then, for example, a lift vs.time graph should first steep up rather quickly (the cascade phase), and then even out.

So, this is basically my extended answer to "why do airplanes fly"...

 

[PBRMEASAP]

Thank you, arildno, for that extremely in depth explanation! It was very clear and well thought out. Now I have a much better picture of what's going on.

I have not yet encountered the counter-spinning vortex explanation, but at least now I'll recognize it when I see it. Since you brought up the subject of pressure in an earlier post, I will take this opportunity to admit my ignorance about it. I understand the concept of pressure in a fluid when viewed in its rest frame--the particles have zero average velocity, but the small fluctuations about zero create an equal pressure in all directions. But it does seem a little counterintuitive to me that there is still no preferred direction even when the fluid is moving with average velocity V. For instance, Thomas2 keeps bringing up the case of "dust" (not an inviscid fluid, as he says), in which there are no interactions between the particles. It certainly seems reasonable that if you get pelted with a stream of dust, momentum will be transferred to you, even though there is no "pressure field". Of course, this is mostly because dust doesn't form a coherent body that can flow around you. My question is, how does an inviscid fluid manage to completely avoid this momentum transfer, as in the regular form of d'Alembert's paradox. I know that the resolution to this is that all real fluids are slightly viscuous, and that the momentum gets transferred through the boundary layer. But even in the idealized non viscous case, how does the fluid manage to flow past an obstacle without bouncing off it and losing some momentum to it? This is not obvious to me.

Oh, by the way, I think the title of that book is actually "A Mathematical Introduction to Fluid Mechanics". I got the first two words backwards. But the authors, Chorin and Marsden, are correct.

 

[Thomas2]

Besides being generally wrong, you are also contradicting yourself. That's not what you said in the piece I quoted.
In fact, if the "shadow" of the airstream made a difference, then a flat-bottom airfoil at -4* aoa, or better yet, an airfoil with a large camber (concave underside), would produce a fair amount of negative lift, considering that the entire bottom is in the "shadow". Once again, what you describe does not fit reality. And your contradictions and permutations make it sound like you are making this stuff up as you go along. Not only have you not bothered to learn how it really works, you haven't even thought through your own idea.

You should have a better look at the airfoil profiles. Take for instance http://www.netax.sk/hexoft/stunt/images/342.gif (which is from the page you quoted yourself in the thread https://www.physicsforums.com/showthread.php?t=66840&page=4&pp=15 recently (post #55)) : the highest point of the camber both at the top and bottom is towards the left (upstream) of the center, i.e. both the upper and lower side should produce a positive lift here. It is in fact the normal convex underside that should produce a negative lift, but since the curvature is less than for the upper side the resultant lift is then still positive.

P.S.: 'Shadow' is defined here as those parts of the surface where the normal has a component parallel to the airstream rather than anti-parallel. Hence the parts of the lower surface to the right (downstream) of the camber maximum are not in the shadow for the concave underside.

 

[arildno]

PBRMEASAP:
D'Alembert's paradox ONLY appears for a body who moves with constant velocity in an inertial frame (whose rest frame is thereby also a rest frame) when the motion of an unbounded fluid about it is stationary with respect to the body's rest frame.

If the body is accelerating, or the motion of the fluid cannot be regarded as stationary within the body's rest frame, then there are certainly forces predicted to work on the body.
In many cases, those predicted forces may well swamp the also present frictional forces; i.e, inviscid theory predicts accurately.

Effectively, it boils down to what is the equilibrium pressure distribution the body will provoke the fluid to generate/tend to?
It so happens, that that equilibrium distribution for an inviscid fluid (with no initial circulation) instantiates D'Alembert's paradox.

Thus we have to clearly distinguish between transient phenomena and equilibrium phenomena we tend to achieve; what is present in transient phenomena is not at all necessarily indicative of the proper equilibrium situation.

 

[PBRMEASAP]

D'Alembert's paradox ONLY appears for a body who moves with constant velocity in an inertial frame (whose rest frame is thereby also a rest frame) when the motion of an unbounded fluid about it is stationary with respect to the body's rest frame.

Right, I understand that. I neglected to specify "in an inertial frame" in my last post. I should have said this: it is not immediately clear to me how a body moving through an inviscid fluid reaches a nonzero equilibrium velocity. I know that it does--I'm not disputing simple experimental results. I'm just having a hard time seeing how this occurs. As you said, the fluid adjusts its pressure distribution until there is no net force on the body, and at that point the body is moving with constant velocity. In the case of dust (no interactions between particles), this would not happen since there is no pressure distribution. So clearly a fluid, even an inviscid one, is much more special than dust. My problem is in seeing how to arrive at this result from a typical description of pressure, i.e. that it is the result of random motions of the individual fluid particles.

 

[arildno]

" As you said, the fluid adjusts its pressure distribution until there is no net force on the body.." FROM THE FLUID!

Oh, you need a thrust force (say, from an engine) to get the body moving in the first place!
As long as we're in the time-dependent phase, you'll need a non-zero thrust force to oppose the drag in the inviscid fluid if you want the body to move with CONSTANT velocity.

The pathologies of D'Alembert's paradox then tell us that in order to keep the constant velocity, you may over time reduce your thrust force to zero.

A body which suddenly start moving in an inviscid fluid certainly experience a drag force from the fluid, i.e, you need a thrust force acting on the body to keep it going (or, as you might say, the engine imparts energy to the fluid).
Obviously, this means that the level of "circulation" is only directly related to the force once stationary conditions has set in (since the circulation level remains constant throughout time). Before that happens, there isn't any connection.

HMM..I'm not altogether certain I've answered your question.

 

[PBRMEASAP]

Well, not quite, but you are helping me clarify my question . In an inviscid fluid, after all transient motion has died down, a body can move with a finite velocity without needing thrust to keep it going. But this doesn't happen for a body moving through dust. After infinite time, the body comes to rest. This discrepancy must have something to do with pressure. So far, all I know about pressure is that it has something to do with the random motion of individual fluid particles. Is there more to it than that? How does the random motion of the fluid particles allow this quite remarkable thing to happen? I'm trying to get a more complete picture of what fluid pressure is.

 

[arildno]

Very good question!

You highlight a subtlety about the mathematical, ideal fluid which I certainly should try to answer:

Given a homogenous, incompressible, inviscid fluid there is no mechanism present for dissipation of kinetic energy!
Nor can the large scale fluid motion be coupled to temperature changes indirectly through a thermodynamic state relation, since the density is constant for a homogeneous, incompressible medium.
Thus, the equations of motion and mass conservation forms a CLOSED system on their own, with pressure and velocities as our unknowns; i.e, the concept&reality of temperature is wholly ignored here.


That is, we have no actual heat production in our unbounded domain; so whatever (macroscopic) kinetic energy comes in by aid of the engine must remain there in the form of (macroscopic) kinetic energy for all time, since the domain doesn't have any boundaries through which the kinetic energy can escape..
(Alternatively, referring to the Bernoulli equation, we can say that energy will be stored either only in the forms of kinetic energy or pressure; i.e, pressure might be regarded as a sort of potential energy)

In the real case, the body&the fluid come to rest somewhat heated.

Thus, the inviscid fluid has a somewhat twisted picture of pressure as well:
It models correctly how pressure is force per area acting strictly normal onto a surface, and that it will provide a macroscopic acceleration along its negative gradient.
Beyond that correct facet however, the inviscid fluid model completely ignores pressure's connection to the actual, random thermal motion of molecules.

That is, we have a mathematical model which captures what is often the main macroscopic dynamics ("pure" pressure" dynamics)
It is a clever approximation to reality, that's all.

The value of an appximation is (at least) two-fold:
1. It is in general simpler to solve than the "real" problem; hence, when we may expect to yield accurate results, we don't waste a lot of time trying to solve the real, intractable problem.

2. By studying special cases, particularly when the model fails gloriousy (as this one does with D'Alembert's paradox), this provides us with a clue as to what counter-acting mechanisms nature uses which we didn't take into account in our simple model.
But such cases also signify the trends to which nature would tend without that opposing mechanism, i.e, we deepen our understanding of the dynamics in the one facet we chose to include.

 

[Thomas2]

Aerodynamic Lift vs. Magnus Effect

I think it is important in this context to point out the fundamental difference between the aerodynamic lift and the magnus effect. As indicated in my posts #66 and #70 (page 5), the former should exist also for a strictly non-viscous gas, but using the same argumentation as there, the magnus effect does not.
Consider a rotating ball that is moving through an inviscid gas (i.e. molecules interacting with the ball but not with each other): if the surface of the ball would be mathematically smooth, then the rotation would actually be without any effect at all because the air molecules would just bounce off like for a non-rotating sphere, but even for a realistic rough surface (obviously a surface can not be smoother than about 1 atomic radius), the overall effect still cancels to zero: the pressure on the side rotating against the airstream is higher at the front but smaller at the back (and the other way around for the co-rotating side) so overall there is no resultant force on the ball but merely a torque that slows down the rotation.
Hydrodynamics arguments (i.e. Bernoulli's principle) are therefore required to explain the magnus effect but not for the usual aerodynamic lift.

 

[PBRMEASAP]

Thus, the inviscid fluid has a somewhat twisted picture of pressure as well:
It models correctly how pressure is force per area acting strictly normal onto a surface, and that it will provide a macroscopic acceleration along its negative gradient.
Beyond that correct facet however, the inviscid fluid model completely ignores pressure's connection to the actual, random thermal motion of molecules.

Thank you for clearing that up! I had suspected this was the case, but I didn't have an argument to back it up.

 

[arildno]

Thank you for clearing that up! I had suspected this was the case, but I didn't have an argument to back it up.

OK, then I'm just about finished (unless you have some other questions).

 

[PBRMEASAP]

Yes. Apparently he only reads his own posts and no one else's.

edit: I think you've answered them all. Thanks again! (Didn't want to make a separate post)

 

[arildno]

Yes. Apparently he only reads his own posts and no one else's.

edit: I think you've answered them all. Thanks again! (Didn't want to make a separate post)

It has been a pleasure!

 

[russ_watters]

You should have a better look at the airfoil profiles. Take for instance http://www.netax.sk/hexoft/stunt/images/342.gif (which is from the page you quoted yourself in the thread https://www.physicsforums.com/showthread.php?t=66840&page=4&pp=15 recently (post #55)) : the highest point of the camber both at the top and bottom is towards the left (upstream) of the center, i.e. both the upper and lower side should produce a positive lift here. It is in fact the normal convex underside that should produce a negative lift, but since the curvature is less than for the upper side the resultant lift is then still positive.

You're changing your claims, but in any case, they still conflict with what is actually observed to occur: All cambered airfoils, regardless of where the point of max thickness occurs and regardless of if the bottom is flat, convex, or concave, produce lift at negative aoa.

You do know about LAMINAR FLOW AIRFOILS, right? These airfoils have the point of maximum thickness further back than in typical airfoils - up to 50% of the way back. Yet, they are more efficient than typical airfoils (the reason they are not used is due to flow stability, not efficiency).

What I don't get is why you don't actually test this yourself. Its relatively simple (I gave a link where a guy built a crude wind tunnel in his house, but you could also build a little model and throw it...) and you'd save yourself from being so spectacularly wrong all the time. Build a little model or wind tunnel, test it, then flip the airfoil around backwards and test again.

edit: HERE are some laminar flow airfoils (on hydrofoils, but the principle is the same) with max thickness 50% of the way back - they produce lift at zero aoa.

 

[Doc Al]

I think its time to put a lid on this thread. Some great answers here, that I'm still digesting. (Good job arildno and others!)

 

[Doc Al]

Locking of the stagnation point

arildno requested that I add this to the thread, as he didn't want to leave any gaps in his explanation:

From arildno:

Unfortunately, I glossed over a relevant topic because it is "too obvious", but on further reflection, I've found that my argument on why a viscous fluid favours downwash really becomes untenable without broaching it.
It concerns the "trivial" fact that for a viscous fluid, a stagnation point becomes locked onto the leading edge.
Clearly, that high pressure zone will give a fluid particle somewhat above the wing a horizontal acceleration component away from the leading stagnation point.
Thus, that fluid particle does not only, as I seemed to suggest, get a roughly normal acceleration onto the wing, but also a tangential acceleration down the wing, providing its "punch".

Locking the stagnation point on the leading edge effectively replaces the unphysical mechanism through which an ideal fluid effects tangential downrush:
It places its stagnation point on the downside of the wing, fluid rush up towards the leading edge, twists about, and rush downwards the upper side.
On the other side of the stagnation point on the underside, the fluid rush down to the trailing edge, twist about it, and the backflow then rush up to meet the downrush in a new stagnation point.
I.e, in the D'Alembert case, we have infinite suction pressure at BOTH edges..

Thus, the leading edge behavior in a real fluid is to replace a totally unphysical mechanism for downrush on the upper side with the mechanism of the frontal stagnation pressure, whereas viscosity's role at the trailing is to reduce upflow.

Both these mechanisms are succincntly described in russ waters' first link, i.e, that viscosity tends to DAMPEN velocity gradients.​